Mathematician Finds Easier Way to Solve Quadratic Equations

[Albert Chan]

"Easier" way is really for educationally purpose, instead of memorizing quadratic formula.

Let roots = m ± g (all parabola look like this)

(x - (m+g)) * (x - (m-g)) = x² + (-2m) x + (m²-g²)

Match this to x² + (b/a) x + (c/a) = 0

b/a = -2m
c/a = m²-g²

LHS is just constants. Solve for m, then for g

We can use same pattern trick for cubic roots. Solve for y, then z = p/(-3y)

(07-20-2021 10:12 PM)Albert Chan Wrote:  x³ + y³ + z³ − 3xyz = (x+y+z) * (x+yω+z/ω) * (x+y/ω+zω), where ω = e^(i*2*pi/3)

LHS is a depressed cubic: x³ + p*x + q , where p = -3yz, q = y³ + z³

We then setup a quadratic of t, with roots (y³, z³)

(t - y³)*(t - z³) = t² - q*t - (p/3)³       → t = q/2 ± √Δ, where Δ = (q/2)² + (p/3)³

https://math.stackexchange.com/questions...2ab-omega2